Question

Question #0973a

Answer 1

`ln(2+x) = ln2 + sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))`

Explanation:

Write the function as:

`ln(2+x) = ln (2 (1+x/2)) = ln2 +ln(1+x/2)`

Now use the MacLaurin development of `ln (1+x)`, substituting `x/2` to `x`:

`ln(1+x/2) = sum_(n=0)^oo (-1)^n (x/2)^(n+1)/(n+1) = sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))`

In conclusion:

`ln(2+x) = ln2 + sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))`